Understanding a Wattmeter 1

[ubern]

We have a Wattmeter that uses the Hall effect to take the instantaneous value of voltage and current and sends it to a DC meter. What im wondering is if this is actually reading KW or just a close approximation.

P = 3^.5 * I * V * pf. So if we have 2 generators in parallel and adjust the one of the voltage regulators down and the other one up then P will be the same and V will be the same. So sense pf changes then current must change. So on a plot of the sine waves for voltage and current on, the generator that i raised voltage on, i would see the phase angle increase AND the amplitude of the current wave.

So doesnt this mean that pf is affecting the instantaneous value of voltage and current and our Wattmeter is only reading something close to real power?

 

[Skogsgurra]

Rest assured that the principles are correct and that you get the true kW and kWh readings.

The instantaneous multplication of current and voltage takes place typically around 100 times per period. The results are then integrated to form the correct sum. This technique makes the modern kWh-meter just as reliable and unsensitive to distortion and power factor variations as were the old electromechanical meters.

I remember you putting a similar question only some week ago. Is this a big problem in your operation?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

[red]"just as unsensitive to distortion"[/red]

Actually the new ones are much better at correctly measuring distorted waveforms.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ubern]

No, we are not having any equipment problems its just a level of knowledge thing. See we train on how to operate our equipment but very little on how it works. That last question about the reverse power relay was a real eye opener. Once I understood that pf could be represented on a graph with x axis being motoring/generating and y axis being leading/lagging, a lot of the things i had observed but couldnt explain started to make sense.

I still don't see how it could give an accurate reading. Can you please explain what is happening in a little more detail? Our meters are not digital, the field in the transducer is provided the current signal applied to coils on either side of a crystal. the potential signal is applied to the crystal and the DC voltage that is generated at 90 degrees to the potential signal is sent to an analog DC meter.

Am i miss-understanding what is ment by instantaneous value? I understand it as: pick a point on the voltage sine wave and the current value would be whatever point had the same X axis value..........Heres what i dont get. lets say you make the generator more lagging...then the current value for a given X will go down. Sense the amplitude of the sine wave also goes up in this situation then the difference in value between potential and current would go back up but because the current sine wave is distorted when amplitude is raised then it would not return the difference in values back to original.

Thank you for your help =)

 

[waross]

If you increase the voltage level of one generator of a pair, the current will increase also. BUT, the phase angle between the voltage and the current will change.
Now, if you draw a current waveform and a voltage wave form on the same graph and look at instantaneous values of both current and voltage, you will see that at times the voltage will be negative while the current is positive, and several other instances of the voltage and current having opposite signs. (+ or -)
In these areas the product of volts times amps is negative and subtracts from the total.
When a generator is sharing power with other sources, the basic formula, (power in = power out plus losses) holds true.
The power delivered by a paralleled generator depends on the throttle setting.
I think I will say that again.
The power delivered by a paralleled generator depends on the throttle setting.
When you play with the excitation, you are shifting the angle between the voltage and the current. You are not changing the power delivered. The instantaneous, simultaneous sampleing of both the current and the voltage by the electronic meter detects the negative power periods and subtracts them from the total.

Bill
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"Why not the best?"
Jimmy Carter

 

[Skogsgurra]

I made a little Excel sheet to demonstrate was Bill was just saying about phase angle between voltage and current. It is also possible to demonstrate that harmonics do not produce any net power. See attached file.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...